Referring to the AP Inter 2nd Year Maths Study Material Chapter 8 Application of Integrals Exercise 8a Solutions makes it easier to understand complex problems.
AP Inter 2nd Year Maths Application of Integrals Solutions Exercise 8a
I.
Question 1.
Find the area of the region bounded by the ellipse \(\frac{x^2}{16}+\frac{y^2}{9}=1\)
Solution:
We have \(\frac{x^2}{16}+\frac{y^2}{9}=1 \Rightarrow \frac{y^2}{9}=1-\frac{x^2}{16} \Rightarrow y^2=9\left(\frac{16-x^2}{16}\right) \Rightarrow y=\frac{3}{4} \sqrt{16-x^2}\)
Ares of the ellipse = 4 × Area(OAB)
A = \(4 \int_0^4 \mathrm{ydx}=4\left(\frac{3}{4}\right) \int_0^4 \sqrt{16-\mathrm{x}^2} \mathrm{dx}\)
= \(3\left[\frac{x}{2} \sqrt{16-x^2}+\frac{16}{2} \sin ^{-1} \frac{x}{4}\right]_0^4=3\left[2 \sqrt{16-16}+8 \sin ^{-1}(1)-0-8 \sin ^{-1}(0)\right]\)
= 3\(\left[\frac{8 \pi}{2}\right]\) = 3[4π]= 12π
∴ Ares of the given ellipse = 12π sq.units
Question 2.
Find the area of the region bounded by the ellipse \(\frac{x^2}{4}+\frac{y^2}{9}=1\)
Solution:
We have \(\frac{x^2}{4}+\frac{y^2}{9}=1 \Rightarrow \frac{y^2}{9}=1-\frac{x^2}{4} \Rightarrow y^2=9\left(\frac{4-x^2}{4}\right) \Rightarrow y=\frac{3}{2} \sqrt{4-x^2}\)
Area of the ellipse A = 4 × Area(OAB)
A = \(4 \int_0^2 \mathrm{ydx}=4\left(\frac{3}{2}\right) \int_0^2 \sqrt{4-\mathrm{x}^2} \mathrm{dx}\)
= \(\left[\frac{x}{2} \sqrt{4-x^2}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_0^2=6\left[\frac{2 \pi}{2}\right]=6 \pi\)
∴ Ares of the given ellipse = 6π sq.units